A condenser is charged to a potential difference of 120V. It's energy is 1×10−5J. If battery is there and the space between plates is filled up with a dielectric medium (εr=5). Its new energy is
5×10−5J
10−5J
3×10−5J
2×10−5J
A
10−5J
B
3×10−5J
C
5×10−5J
D
2×10−5J
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Solution
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Given energy = 1×10−5 When battery is connected, V across capacitor will be constant. When dielectric is filled, C increases by K ∴Newenergy=12(KC)V2 =K12CV2(Given:12CV2=1×10−5J) =K(1×10−5)J =5×(1×10−5)J =5×10−5J
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